Closed Subset of Real Numbers with Lower Bound contains Infimum
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Theorem
Consider the real number line as a metric space under the usual metric.
Let $A \subseteq \R$ such that $A$ is closed in $\R$ and $A \ne \O$.
Let $A$ be bounded below.
Then $A$ contains its infimum.
Proof
From Infimum of Bounded Below Set of Reals is in Closure:
- $\inf A \in \map \cl A$
From Set is Closed iff Equals Topological Closure:
- $A = \map \cl A$
Therefore $\inf A \in A$.
$\blacksquare$
Sources
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Exercise $5$